What is the relative minimum, relative maximum, and points of inflection of $f (x) = x^{4} - 4 x^{2}$ $f(x) = x^4 - 4x^2$ ?

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Answer 1 · 2016-11-24T20:41:54

$f'(x)=4x^3-8x$
$f'(x)=4x(x^2-2)$

$4x(x^2-2)=0$

$x=0, x=+-sqrt2$

The multiplicity of each of these zeros is odd, therefore, there will be a minimum or maximum at each value.

For:
$x>sqrt2$
$y>0$
Positive slope

For:
$0 < x < sqrt2$
$y < 0$
Negative slope

For:
$-sqrt2 < x < 0$
$y > 0$
Positive slope

For :
$x < -sqrt2$
$y < 0$
Negative slope

$:. x=sqrt2 =>$ Local minimum
$:. x=0 =>$ Local maximum
$:. x=-sqrt2 =>$ Local minimum

$f''(x)=12x^2-8$
$f''(x)=4(3x^2-2)$

$3x^2-2=0$
$x=+-sqrt(2/3)$

For:
$x>sqrt(2/3)$
$y>0$
Concave up

For:
$-sqrt(2/3) < x < sqrt(2/3)$
$y<0$
Concave down

For:
$x < -sqrt(2/3)$
$y>0$
Concave up

$:.$ There are points of inflection at:
$x=+-sqrt(2/3)$

What is the relative minimum, relative maximum, and points of inflection of f(x) = x^4 - 4x^2f(x)=x4−4x2f(x) = x^4 - 4x^2?